What is the integral from-1 to 1 of (x^2+x+1) ?

The integral from-1 to 1 of (x^2+x+1) is 8/3

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integral from-1 to 1 of (x^2+x+1)

\int_{- 1}^{1} (x^{2} + x + 1) d x

\frac{8}{3}

Decimal

2.66666 \dots

Solution steps

\int_{- 1}^{1} x^{2} + x + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{1} x^{2} d x + \int_{- 1}^{1} x d x + \int_{- 1}^{1} 1 d x

\int_{- 1}^{1} x^{2} d x = \frac{2}{3}

\int_{- 1}^{1} x d x = 0

\int_{- 1}^{1} 1 d x = 2

= \frac{2}{3} + 0 + 2

Simplify

\frac{2}{3} + 0 + 2 : \frac{8}{3}

= \frac{8}{3}

What is the integral from-1 to 1 of (x^2+x+1) ?
The integral from-1 to 1 of (x^2+x+1) is 8/3